In the previous section, we saw examples on limits of polynomial functions and rational functions. In this section, we will see the second theorem.
Theorem 2
For any positive integer n,
limx→a[xn−anx−a] = nan−1
• The expression in this theorem is true even if n is any rational number and a is positive.
• We will see the proof for this theorem in higher classes. At present we will see some solved examples.
Solved example 13.3
(i) Evaluate limx→a[x15−1x10−1]
(ii) Evaluate limx→0[√1+x−1x]
Solution:
Part (i):
We have:
limx→1[x15−1x10−1] = limx→1[x15−1x−1 ÷ x10−1x−1] = limx→1[x15−1x−1] ÷ limx→1[x10−1x−1] = [15×115−1] ÷ [10×110−1] = [15×114] ÷ [10×19] = [15×1] ÷ [10×1] = 1510 = 32
Part (ii):
1. Let u = x +1
2. We want the limit when x approaches zero.
• Based on (1), we can write:
When x approaches zero, u will approach 1
• Also based on (1), we get: x = u-1
3. So the required limit can be written in another form:
limu→1[√u−1u−1]
4. Now we can evaluate the limit:
limx→0[√1+x−1x] = limu→1[√u−1u−1] = limu→1[u12−1u−1] = 12×1(12−1) = 12×1(−12) = 12×1 = 12
Alternate method:
limx→0[√1+x−1x] = limx→0[√1+x−1x×√1+x+1√1+x+1] = limx→0[1+x−1x(√1+x+1)] = limx→0[xx(√1+x+1)] = limx→0[1(√1+x+1)] = [1(√1+0+1)] = [1(1+1)] = 12
◼ Remarks:
• In the first line, we multiply both numerator and denominator by "√(1+x) + 1".
♦ So the numerator will be in the form (a+b)(a-b)
♦ We know that, (a+b)(a-b) = a2 - b2.
In the next section, we will see limits of trigonometric functions.
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